SOLUTION: Meg rowed her boat upstream at a distance of 48 mi. then rowed back to the starting point. The total time of the trip was 16 hours. If the rate of the current was 4mph, find the a

Question 1132693: Meg rowed her boat upstream at a distance of 48 mi. then rowed back to the starting point. The total time of the trip was 16 hours. If the rate of the current was 4mph, find the average speed of the boat relative to the water
Found 3 solutions by stanbon, ikleyn, greenestamps:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Meg rowed her boat upstream at a distance of 48 mi. then rowed back to the starting point. The total time of the trip was 16 hours. If the rate of the current was 4mph, find the average speed of the boat relative to the water
-----
Upstream DATA::
dist = 48 mi ; time = t hours ; rate = 48/t mph
---
Downstream DATA:
dist = 48 mi ; time = 16-t hrs ; rate = 48/(16-t) mph
----
Rate equations:
boat + 4 mph = 48/l6-t)
boat - 4 mph = 48/t
-----
Solve for "boat":
2*boat = 48/(16-t) + 48/t
-----
boat rate in still water = = 24/(16-t)+ 24/t = [24t+24*16-24t]/[t(16-t)]
= [24*16]/[16t-t^2]
----

Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.

Let "x" be the average speed of the boat relative to the water (usually called as a "boat speed in still water"), in miles per hour. Then the speed of the boat downstream is (x+4) mph, and the time traveling 48 miles downstream is hours. The speed of the boat upstream is (x-4) mph, and the time traveling 48 miles upstream is hours. Total time for the round trip is 16 hours (given !), which gives you the "time" equation + = 16 hours. (1) It is your basic equation to solve. At this point, I just know the answer: it is 8 miles per hour for the boat speed in still water. But you, probably, want to get the solution on formal algebra way. OK, let's do it. Multiply equation (1) by (x-4)*(x+4) = . You will get 48*(x+4) + 48*(x-4) = 16*(x^2-16). Simplify it step by step. 48x + 4*48 + 48x - 4*48 = = 0. Factor out the common factor of 16 and cancel it. You will get = 0. Factor left side (x-8)*(x+2) = 0. The two roots of the last equation are x= 8 and x= -2. In this problem only positive root x= 8 is meaningful solution. So, we got the same answer as I anticipated above. Answer. The boat speed in still water is 8 miles per hour. CHECK. Let's check equation (1). Its left side is + = + = 12 + 4 = 16 hours - same as the given total time. ! The solution is correct !

Solved.

------------------

It is very standard and typical round trip upstream and downstream Travel and Distance problem.

See the lessons
    - Wind and Current problems
    - More problems on upstream and downstream round trips
    - Wind and Current problems solvable by quadratic equations
    - Unpowered raft floating downstream along a river
    - Selected problems from the archive on the boat floating Upstream and Downstream
in this site, where you will find other similar solved problems with detailed explanations.

Read them attentively and learn how to solve this type of problems once and for all.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the section "Word problems", the topic "Travel and Distance problems".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

/\/\/\/\/\/\/\/\/\/\/\/\/

I read the "solution" by @stanbon and found it totally wrong.

For your safety, simply ignore it.

It is why I presented my solution here.

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

The formal algebraic solution by tutor @ikleyn is fine, and well presented.

However, with the information given, this problem can be solved mentally with a bit of trial and error.

Since the rate of the current is 4mph, the difference between the upstream and downstream speeds will be 8mph.

Then, since the total time for the trip was EXACTLY 16 hours, the speed of the boat in still water will almost certainly be a whole number; and then the times for the upstream and downstream trips will be whole numbers.

A very small bit of mental arithmetic finds that two speeds (mph) that differ by 8 and are factors of 48 (miles) are 4 and 12.

And indeed these speeds work with the given information -- the total time for the trip is 48/4+48/12 = 12+4 = 16 hours.

So the upstream and downstream rates are 4mph and 12mph; since the rate of the current is 4mph, the rate of the boat in still water is 8mph.
RELATED QUESTIONS
Meg rowed her boat upstream a distance of 39 mi and then rowed back to the starting... (answered by stanbon)
Meg rowed her boat upstream a distance of 91 mi and then rowed back to the starting... (answered by mananth)
Meg rowed her boat upstream a distance of 15 mi and then rowed back to the starting... (answered by ikleyn)
Meg rowed her boat upstream a distance of 65 mi and then rowed back to the starting... (answered by ikleyn,greenestamps)
Meg rowed her boat upstream a distance of 5 miles and then rowed back to the starting... (answered by mananth)
Meg rowed her boat upstream a distance of 5 miles and then rowed back to the starting... (answered by mananth)
Meg rowed her boat upstream a distance of 72 miles and then rowed back to the starting... (answered by josgarithmetic)
Meg rowed her boat upstream a distance of 40 miles and then rowed back to the starting... (answered by mananth)
Meg rowed her boat upstream a distance of 21 miles and then rowed back to the starting... (answered by richwmiller)